How to filter out any digit that contains odd number in a range of number in Python

Question

I want to filter out any number that contains a odd digit in it, this should happen in a range of number.

e.g.

1-10 must give an output as follows:

2 4 6 8

Answer 1

This code will work:

scsv = ""
for i in range (1000,3001):
    if i % 2 == 0:
        if '1' not in str(i) and '3' not in str(i) and '5' not in str(i) and '7' not in str(i) and '9' not in str(i):
             scsv = scsv+str(i)+" , "
print (scsv[:-3])

This code is very basic but serves the purpose.

Answer 2

n = list(range(10))

b = list(filter(lambda i:i%2!=0,n))

print(b)